The Interior of Sets in Finite Topological Products

In the following theorem we will show that if $\{ X_1, X_2, ..., X_n \}$ is a finite collection of topological spaces and $A_i \subseteq X_i$ for all $i \in \{ 1, 2, ..., n \}$ then the interior of the product of these sets is equal to the product of the interiors of these sets.

Theorem 1: Let $\{ X_1, X_2, ..., X_n \}$ be a collection of topological spaces and let $A_i \subseteq X_i$ for all $i \in \{1, 2, ..., n \}$. Then $\displaystyle{\mathrm{int} \left ( \prod_{i=1}^{n} A_i \right) = \prod_{i=1}^{n} \mathrm{int} (A_i)}$.

• Proof: Let $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n) \in \mathrm{int} \left ( \prod_{i=1}^{n} A_i \right )}$. Then there exists an open set $\displaystyle{U = \prod_{i=1}^{n} U_i}$ in $\displaystyle{\prod_{i=1}^{n} X_i}$ such that:

(1)

\begin{align} \quad \mathbf{x} \in U = \prod_{i=1}^{n} U_i \subseteq \prod_{i=1}^{n} A_i \end{align}

• So then for all $i \in \{1, 2, ..., n \}$ we have that $x_i \in U_i \subseteq X_i$. Each of the open sets $U_i$ is open in $X_i$, and so $x_i \in \mathrm{int}(X_i)$ for all $i \in \{1, 2, ..., n \}$. Thus $\displaystyle{\mathbf{x} \in \prod_{i=1}^{n} \mathrm{int} (A_i)}$ which shows that:

(2)

\begin{align} \quad \mathrm{int} \left ( \prod_{i=1}^{n} A_i \right) \subseteq \prod_{i=1}^{n} \mathrm{int} (A_i) \quad (*) \end{align}

• Now let $\displaystyle{\mathbf{x} = (x_1, x_2, ..., x_n) \in \prod_{i=1}^{n} \mathrm{int} (A_i)}$. Then $x_i \in \mathrm{int} (A_i)$ for all $i \in \{1, 2, ..., n \}$. So for each $i$ there exists an open set $U_i$ of $X_i$ such that:

(3)

\begin{align} \quad x_i \in U_i \subseteq X_i \end{align}

• Let $\displaystyle{U = \prod_{i=1}^{n} U_i}$. Then $U$ is open in $\displaystyle{\prod_{i=1}^{n} X_i}$ as it is a product of open sets. Moreover:

(4)

\begin{align} \quad \mathbf{x} \in U = \prod_{i=1}^{n} U_i \subseteq \prod_{i=1}^{n} A_i \end{align}

• Thus $\displaystyle{\mathbf{x} \in \mathrm{int} \left ( \prod_{i=1}^{n} A_i \right )}$ which shows that:

(5)

\begin{align} \quad \mathrm{int} \left ( \prod_{i=1}^{n} A_i \right) \supseteq \prod_{i=1}^{n} \mathrm{int} (A_i) \quad (**) \end{align}

• From the inclusions in $(*)$ and $(**)$ we conclude that:

(6)

\begin{align} \quad \mathrm{int} \left ( \prod_{i=1}^{n} A_i \right) =\prod_{i=1}^{n} \mathrm{int} (A_i) \quad \blacksquare \end{align}